Find Maximum Subarray Divide and Conquer in DSA Go - Time & Space Complexity

Time Complexity: Find Maximum Subarray Divide and Conquer

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum subarray grows as the input array gets bigger.

Specifically, how does the divide and conquer method affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func maxSubArray(arr []int, low, high int) int {
  if low == high {
    return arr[low]
  }
  mid := (low + high) / 2
  leftMax := maxSubArray(arr, low, mid)
  rightMax := maxSubArray(arr, mid+1, high)
  crossMax := maxCrossingSubArray(arr, low, mid, high)
  return max(leftMax, rightMax, crossMax)
}

func maxCrossingSubArray(arr []int, low, mid, high int) int {
  leftSum := -1 << 31
  sum := 0
  for i := mid; i >= low; i-- {
    sum += arr[i]
    if sum > leftSum {
      leftSum = sum
    }
  }
  rightSum := -1 << 31
  sum = 0
  for j := mid + 1; j <= high; j++ {
    sum += arr[j]
    if sum > rightSum {
      rightSum = sum
    }
  }
  return leftSum + rightSum
}
    

This code finds the maximum sum of a subarray by splitting the array, solving for left, right, and crossing parts, then combining results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls splitting the array into halves.
• How many times: The array is split until single elements remain, about log n levels.
• Additional operation: At each level, the crossing subarray is found by scanning left and right parts once (linear scan).

How Execution Grows With Input

At each level, the work to find the crossing max is proportional to the size of the current subarray.

Input Size (n)	Approx. Operations
10	About 10 * log 10 ≈ 33
100	About 100 * log 100 ≈ 664
1000	About 1000 * log 1000 ≈ 9966

Pattern observation: As input size grows, total work grows a bit more than linearly, roughly n times log n.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the size of the input, because we do linear work at each level of a logarithmic number of levels.

Common Mistake

[X] Wrong: "Since the array is split in half each time, the time complexity is O(log n)."

[OK] Correct: Although the array splits reduce problem size, at each level we still scan parts of the array linearly, so total work adds up to more than just log n.

Interview Connect

Understanding this time complexity helps you explain how divide and conquer can speed up problems compared to simple loops, showing your grasp of algorithm design.

Self-Check

"What if we replaced the crossing subarray scan with a constant time operation? How would the time complexity change?"

Practice

1. What is the main idea behind the Divide and Conquer approach to find the maximum subarray sum?

easy

A. Split the array into two halves, find max subarray in left, right, and crossing middle, then combine results

B. Sort the array and pick the largest element as max subarray sum

C. Use a sliding window to find the max sum subarray in one pass

D. Calculate sum of all subarrays and pick the maximum

Solution

1. Step 1: Understand divide and conquer strategy

   The array is split into two halves recursively to simplify the problem.

2. Step 2: Combine results from subproblems

   Find max subarray in left half, right half, and crossing the middle, then take the maximum of these three.

3. Final Answer:

   Split array, find max in left, right, crossing middle, combine -> Option A

4. Quick Check:

   Divide and conquer = split and combine max subarrays [OK]

Hint: Divide array, conquer halves, combine max sums [OK]

Common Mistakes:

• Thinking sorting helps find max subarray sum
• Confusing divide and conquer with sliding window
• Calculating all subarrays sums directly

2. Which of the following Go function signatures correctly represents the divide and conquer function to find maximum subarray sum in an array arr between indices low and high?

easy

A. func maxSubArray(arr []int, low int high int) int

B. func maxSubArray(arr int[], low int, high int) int

C. func maxSubArray(arr []int, low int, high int) (int, int)

D. func maxSubArray(arr []int, low, high int) int

Solution

1. Step 1: Check Go slice and parameter syntax

   In Go, slices are declared as []int, and multiple parameters of same type can be grouped.

2. Step 2: Identify correct function signature

   func maxSubArray(arr []int, low, high int) int uses correct slice type and groups low, high as int parameters, returning int sum.

3. Final Answer:

   func maxSubArray(arr []int, low, high int) int -> Option D

4. Quick Check:

   Go slice syntax and grouped params = func maxSubArray(arr []int, low, high int) int [OK]

Hint: Use []int for slices and group same-type params [OK]

Common Mistakes:

• Using int[] instead of []int for slices
• Returning multiple values when only sum needed
• Separating same-type parameters unnecessarily

3. What is the output of the following Go code snippet that uses divide and conquer to find maximum subarray sum?

package main
import "fmt"

func maxCrossingSum(arr []int, low, mid, high int) int {
    leftSum := -1 << 31
    sum := 0
    for i := mid; i >= low; i-- {
        sum += arr[i]
        if sum > leftSum {
            leftSum = sum
        }
    }
    rightSum := -1 << 31
    sum = 0
    for j := mid + 1; j <= high; j++ {
        sum += arr[j]
        if sum > rightSum {
            rightSum = sum
        }
    }
    return leftSum + rightSum
}

func maxSubArray(arr []int, low, high int) int {
    if low == high {
        return arr[low]
    }
    mid := (low + high) / 2
    leftMax := maxSubArray(arr, low, mid)
    rightMax := maxSubArray(arr, mid+1, high)
    crossMax := maxCrossingSum(arr, low, mid, high)
    if leftMax >= rightMax && leftMax >= crossMax {
        return leftMax
    } else if rightMax >= leftMax && rightMax >= crossMax {
        return rightMax
    } else {
        return crossMax
    }
}

func main() {
    arr := []int{-2, 1, -3, 4, -1, 2, 1, -5, 4}
    fmt.Println(maxSubArray(arr, 0, len(arr)-1))
}

medium

A. 5

B. 7

C. 6

D. 4

Solution

1. Step 1: Understand the input array and max subarray

   The array is [-2,1,-3,4,-1,2,1,-5,4]. The max subarray is [4,-1,2,1] with sum 6.

2. Step 2: Trace divide and conquer calls

   The function recursively finds max in left, right, and crossing middle, correctly returning 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Max subarray for example array is 6 [OK]

Common Mistakes:

• Choosing sum including -5 which lowers total
• Confusing max crossing sum with max overall
• Miscomputing indices for mid and crossing sums

4. Identify the bug in this Go function for finding max crossing sum and select the fix:

func maxCrossingSum(arr []int, low, mid, high int) int {
    leftSum := 0
    sum := 0
    for i := mid; i >= low; i-- {
        sum += arr[i]
        if sum > leftSum {
            leftSum = sum
        }
    }
    rightSum := 0
    sum = 0
    for j := mid + 1; j <= high; j++ {
        sum += arr[j]
        if sum > rightSum {
            rightSum = sum
        }
    }
    return leftSum + rightSum
}

medium

A. Initialize leftSum and rightSum to very small number (e.g. -1<<31) instead of 0

B. Change loops to run from low to mid and mid+1 to high

C. Return max of leftSum and rightSum instead of sum

D. Add 1 to mid when calculating leftSum loop

Solution

1. Step 1: Analyze initialization of leftSum and rightSum

   They start at 0, which fails if all numbers are negative, missing correct max sums.

2. Step 2: Fix initialization to very small number

   Set leftSum and rightSum to -1<<31 (minimum int) to correctly track max sums even if negative.

3. Final Answer:

   Initialize leftSum and rightSum to very small number (e.g. -1<<31) -> Option A

4. Quick Check:

   Proper initialization avoids missing negative max sums [OK]

Hint: Initialize max sums to very small number, not zero [OK]

Common Mistakes:

• Starting max sums at zero causes wrong results with negatives
• Changing loop directions incorrectly
• Returning sum instead of combined max sums

5. Given an array arr of integers, how would you modify the divide and conquer approach to also return the start and end indices of the maximum subarray along with its sum?

hard

A. Store indices globally and update them only when crossing sum is calculated

B. Change functions to return a struct with sum, start, and end; update recursive calls to track indices

C. Return only sum and find indices by scanning array after result

D. Use a separate loop after divide and conquer to find indices of max sum

Solution

1. Step 1: Modify function return types

   Change functions to return a struct containing sum, start index, and end index to track subarray boundaries.

2. Step 2: Update recursive calls and combine logic

   When combining left, right, and crossing results, also combine their indices accordingly to keep track of max subarray range.

3. Final Answer:

   Return struct with sum, start, end; update recursive calls to track indices -> Option B

4. Quick Check:

   Return sum and indices together for full max subarray info [OK]

Hint: Return sum plus indices in struct, update recursively [OK]

Common Mistakes:

• Using global variables causing bugs in recursion
• Finding indices after sum calculation separately
• Ignoring indices in crossing sum calculation